Answer
The real solutions are: $x_{1}=3$ and $x_{2}=-3$
And the complex solutions are: $x_{3}=i$ and $x_{4}=-i$
Work Step by Step
Given $x^4-8x^2-9=0$
Substituting $u = x^2 ,$ we have:
$x^4-8x^2-9=0 \longrightarrow (x^2)^2-8(x^2)-9=0 \longrightarrow u^2-8u-9=0$
$a = 1, \ b=-8, \ c=-9$
Using the quadratic formula: $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a} ,$ we have:
$\dfrac{-(-8) \pm \sqrt{(-8)^2-4\times 1\times (-9)}}{2\times 1} = \dfrac{8 \pm \sqrt{64+36}}{2} = \dfrac{8 \pm \sqrt{100}}{2} = \dfrac{8 \pm 10}{2} = 4 \pm 5 $
Therefore we have that $u_{1}=4 + 5=9$ and $u_{2}=4-5=-1$
Substituting, we have $(x_{1})^2=u_{1}\longrightarrow (x_{1})^2=9\longrightarrow x_{1}=\sqrt{9}=\pm 3$ and $(x_{2})^2=u_{2}\longrightarrow (x_{2})^2=-1\longrightarrow x_{2}=\sqrt{-1}=\pm i$
